k space function spaces

I ntern.

J.

Math.

&

Math.

Sci.

Vol.

No.

4

(1980)

701-711

701

K-SPACE FUNCTION

SPACES

R.

A. McCOY

Department of Mathematics

Virginia Polytechnic Institute and State University Blacksburg, Virginia 24061 U.S.A.

(Received January 28, 1980)

ABSTRACT. A study is made of the properties on X which characterize when C (X)

is a k-space, where C (X) is the space of real-valued continuous functions on X having the topology of pointwise convergence. Other properties related to the

k-space property are also considered.

KEY WORDS AND

PHRASES.

Function spaces, k-spaces, Sequential spaces,

Fr@chet

spaces, Countable tightness, k-countable,

w-countable.

1980 MATHEMATICS SUBJECT CLASSIFICATION CODES. Primary, 54C35; secondary,

54D50, 54D55, 54D20.

1 INTRODUCTION

If X is a topological space, the notation C(X) is used for the space of all real-valued continuous functions on X. One of the natural topologies on C(X) is the topology of pointwise convergence, where subbasic open sets are those of the

form

for x X and V open in the space of real numbers,

,

with the usual topology.

The space C(X) with the topology of pointwise convergence will be denoted by

c

(x).

For a completely regular space X, C

(X)

is first countable, in fact

metrlz-able, if and only if X is countable

[2].

The purpose of this paper is to show

to what extent this result can be extended to properties more general than first

countability, such as that of being a k-space. Throughout this paper all spaces

will be assumed to be completely regular

Tl-.Spaces.

We first recall the definitions of certain generalizations of first

count-ability. The space X is a

Frchet

space if whenever x A

=

X, there exists a

sequence in A which converges to x. The space X is a s.equentia.l

space

if the open subsets of X are precisely those subsets U such that whenever a sequence

converges to an element of U, the sequence is eventually in U. Also X is a k-space if the closed subsets of X are precisely those subsets A such that for

every

compact’subspace

K X, A N K is closed in K. Finally X has countable

tightness if whenever x A

=

X, there exists a countable subset B

=

A such that x B. The following diagram shows the implications between these properties.

first countable

Frchet

sequential

countable tightness

k-space

We will show that the

Frchet

space, sequential space, and k-space proper-ties are equivalentfor C (X). In order to characterize these properties for C (X) in terms of internal properties of X, we will need to make some additional

definitions. Let (X) be the set of all nonempty finite subsets of X. A

collec-tlon I of open subsets of X is an open cover for finite subsets of X if for every

A (X), there exists a U such that A

=

U. If

[n

is a sequence of

K-SPACE FUNCTION SPACES 703

for every n _{6 IN} (I is the set of natural numbers). In addition, we will say

that

[Un]

is residually

coverin$

if for every x 6 X, there exists an N 6 1 such that for all n N, x 6 U

n

THEOREM I. The following are equivalent.

(a) C (X) is a

Frchet

space. (b) C (X) is a sequential space.

(c) C (X) is a k-space.

(d) Every sequence of open covers for finite subsets of X has a residually

covering string.

PROOF. (d)

=

(a). Suppose that every sequence of open covers for finite

subsets of X has a residually covering string. Let F be a subset of C (X), and let f be an accumulation point of F in C (X). Then for every n q and A

Ix

I

Xk

6 (X), we may choose an

F

x (f(x)

_I

f

+i)

.N

.x

k (f(x

k)

i

f

+i)

f

n,A

I’

1 n

(Xl

n

(Xk

n

1

Also define U(n,A)

[x

6 X

Ifn,A(X)

f(x) <

},

which is an open subset of X.

Then for each n

,

define

n

[U(n’A)

IA6(X)}’

which is an open cover for finite subsets of X. Now

[Un}

has a residually covering string

[U(n,An

so

that for every n IN, we may define f f

n n,A

n

We wish to establish that

If

converges to f in C (X). So let x X, and

n I

let

>

0. There is an N

E

IN with N

-

such that for every n N, x 6

U(n,An).

But then if n >_ N,

i i f (x) f(x)

Ifn,A

(x) f(x)

<

<

n n IN

n

Therefore

{fn(X)}

converges to f(x) for every x 6 X so that

{f

converges to f n

in C (X). Hence C

(X’)

must be a

Fr6chet

space.

(c)

=

(d). Suppose X has a sequence

[n

of open covers for finite subsets

such that no string from

{n

3

is residually covering. Let F

I

41,

and for each

n

> I,

let Ir_n be an open cover for finite subsets of X which refines both V

and

n"

For every n

E

lq and A

E

(X), let U(n,A)

fn

such that A

=

U(n,A), and

let f C(X) be such that f (A)

}

f

(X\U(n,A))

[n

and

n,A n,A n,A

f (X)

[

n].

Then define

n,A

F

If

In

1 and A

(X)}

n,A

and

also

define F*

F\[Co

in C (X), where c is the constant zero function. o

First we establish that F* is not closed in C (X) by showing that c is an

I o

accumulation point of F in C (X). To do this, let

W=

Xl,Vl

N...

O

Xk,Vk

be an arbitrary basic neighborhood of c in C (X). If A

x

_I,

Xk}

and

o n

E

IN such that i V

I N NVk, then fn,A WNF.

We will then obtain that C (X) is not a k-space, as desired, if we can show

that the intersection of F* with each compact subspace of C (X) is closed in that compact subspace. To this end, let K be an arbitrary compact subspace of C (X).

Then for every x X, the orbit

[f(x)

If

K

is bounded in

.

For every x

E

X,

define M(x) sup

[f(x)If

K,

and also for every m q define X

m

Ix

XIM(x)

<

m.

Note that X

[Xmlm

q,

and that for every m,

Xm

c

Xm+

I.

Suppose, by way of contradiction, that for every m, n

E q,

there exists a k n and V

F

k such that Xm c_ V. We define by induction, a string

[Un}

from

[n].

First there exists a k

I i and VI

Fkl

such that XI _c V

I.

For each

i

I,

kl,

choose U

E

so that V

I

=

U Now suppose k and

UI,

,Uk

i i i m

m

have been defined. Then there exists a

km+

I > km

+

i and Vm+l Fk such that

m+l

Xm+

I c_

Vm+

I.

For each i

km

+

I,

km+

I, choose

Ui

i

so that

Vm+

I c U

i.

This defines string

[U ],

which we know to not be residually covering. Let

n

x

E

X be arbitrary. There is an m ₆ such that x 6 X Let n k There is

m m

a j m such that

kj_

_I

+

I

< n <

kj.

Then x

Xm

c

Xj

c

Vj

c

Un.

But this says that

[Un]

is residually covering, which is a contradiction.

We have just established that there exist m, n I such that for every

k >_ n and for every V F_k, X V. Then define M max

[m

n,

let x

E

X be

K-SPACE FUNCTION SPACES 705

i i

arbitrary, and define W

Xo,(-,

)

which is a neighborhood of c in

o

C (X). Suppose f

E

WNF. Then there exists a k

E

lq and A (X) such that

i I

f

fk,A"

Since < f(x

o)

<

,

then k M n. Thus

Xm

U(k,A), so that there exists an x

I

Xm\U(k,A).

But then f(x

I)

k

>

M > m

M(Xl)

so that f K. Therefore

WOFNK

,

so that c is not an accumulation point of F*NK in K.

o

Hence F*NK must be closed in K. Since K was arbitrary, we obtain that

C(X)

is not a k-space.

THEOREM 2. C (X) has countable tightness if and only if every open cover for finite subsets of X has a countable subcover for finite subsets of X.

PROOF. Suppose that every open cover for finite subsets of X has a

count-able subcover for finite subsets of X. Let F be a subset of C (X), and let f be an accumulation point of F in

C(X).

Then for each n and A

Ix

_I

Xk

(X), choose

fn,A

E

F x_I,(f

(Xl)

l’f(Xn

i

+)

N

Xk’

(f

(Xk)

’i

f(x

k)

+)]]

i

Also let U(n,A)

Ix

6 X

If

_n,A(x) f(x) <

}

which is an open subset of X

Then for each n

E

IN,

[U(n,A)

IA

(X)

is an open cover for finite subsets of

X. So for each n 6 IN, there exists a sequence

[A(n,i)

li

] from (X) such

that

[U(n,A(n,i))li

E

is a cover for finite subsets of X. Then define G

{f

In,

i

E

I

n,A(n,i)

To see that f 6 G, let W

[

Xl,Vl

N...

N[

Xk,Vk

be a neighborhood of f in C (X). Let A

Ix

_I,

Xk

and choose n 6 IN so that

(f(xj)

i

n

f(x.) +

I)

c_ _V. for each j i, k. Then there is an i 6 ]q such that

3 n 3

i

fn,A

<n--A c_ U(n,A(n,i)). So for each x A,

(n,i)(x)-

f(x) and hence

f_n,A(n,i) 6W

Conversely, suppose that C (X) has countable tightness, and let N be an open cover for finite subsets of X. For each A 6 (X), let U(A)

E N

be such that A c_ U(A). Also for each n 6 IN and A 6

(X),

let f 6 C(X) be such that

[l,n].

Then define F

f (A)

[

fn

(X\U(A))

In}

and f (X) c_

n,A ,A n,A n

If

In

and A

(X).

n,A

Since the constant zero function, c is an accumulation point of F, then

o

there is a countable subset G of F such that c G. There are sequences

o

[ni

=

q and

[Ai}

=

(X) so that G

If

li

ni,A

i

To see that

[U(Ai)

li

}

is a cover for finite subsets of X, let A

Ix

i,

Xk

(X). Then there exists an i IN such that

fn

i,Ai

xi’

(-i,i)

0 N

x

k,(-l,l)

But this means that A

=

U(Ai),

so that

[U(Ai)

li

IN}

is indeed a cover for finite subsets of X.

Let us now give names to the two properties of X which are expressed in

Theorems 1 and 2. We will call X k-countable whenever C (X) is a k-space, and

we will call X T-countable whenever C (X) has countable tightness. We state

some immediate facts about these properties.

PROPOSITION 3. Every countable space is k-countable.

PROPOSITION 4. Every k-countable space is -countable.

PROPOSITION 5. Every q-countable space is

Lindelf.

PROOF. Let X be T-countable, and let be an open cover of X. Let If be the

family of all finite unions of members of I. Then F is an open cover for finite

subsets of X, so that it has a countable subcover [D for finite subsets of X.

Each member of is a finite union of members of B, so that since

,

covers X,

then I has a countable subcover.

This means that if C (X) has countable tightness, X must be

Lindelf.

In particular,

C(

o)

does not have countable tightness, where

o

is the space of

countable ordinals with the order topology This is in contrast to C (), which

K-SPACE FUNCTION SPACES 707

is an open cover of Xn So for open cover for finite subsets of X, then each

n

Xn

each n

_

IN, has a countable subcollection If such that

{unlu

E

F ] covers

n n

But then

J

[Fnln

E

IN]

is a countable subcollection of I which is a cover for

finite subsets of X. []

COROLLARY 7. Every compact space is q-countable, and every separable metric

space is q-countable.

We now examine some properties of k-countable spaces.

PROPOSITION 8. Every closed subspace of a k-countable space is k-countable. PROOF. Let X be a k-countable space, and let Y be a closed subspace of X.

Let

{Fn

}

be a sequence of open covers for finite subsets of Y. For each n lq,

let U

n

[V

U

(X\Y)

IV

E

fn

}’

which is an open cover for finite subsets

of X. Now

[Un}

has a residually covering string

[Vnj

(X\Y)],

where each

Vn

Fn"

But then

{Vn]

is a residually covering string from

{fn

]"

PROPOSITION 9. Every continuous image of a k-countable space is k-countable.

PROOF. Let X be k-countable, and let f’X Y be a continuous surjection.

Let

{Irn]

be a sequence of open covers for finite subsets of Y. For each n

E

IN, -I

let

n

{f

(V)

IV

fn

]’ which is an open cover for finite subsets of X. Now

-i

[n

] has a residually covering string

{f

(Vn)]’

where each

Vn

fn"

But then

{Vn]

is a residually covering string from

{Vn].

In the next proposition, we use the term covering string, by which we mean

a string which is itself a cover of the space.

PROPOSITION i0. If X is k-countable, then every sequence of open covers of

X has a covering string.

PROOF. Let

[n}

be a sequence of open covers of X. For each n lq, let

f

[U

n.

!fUn+k+

II

kEIN and each U

U

]

n i i

which is an open cover for finite subsets of X. Thus

[Vn]

has a residually covering string

[Vn}.

Now V

I UI

tl...IU

k for some kI IN. Also

Vkl+l

Ukl+l

U.

(JUk2

for some k

2

E

lq with k2

>

k

I.

Continuing by induction, we can

define an increasing sequence

[ki}

such that each V

k

+l=Uk

+IU"

UU

k

i i i+l

This defines

Un

for each n

E

.

To see that

[Un}

is a covering string from

let x X. Then there exists an N such that for all n m N, x V Since n

ki

is increasing, there is some i such that

k.l

N. Then x

Vk.+l

Uk.+IU"

(JUki+l,

so that x is indeed in some

Un.

We next give an important example of a space which is not k-countable.

EXAMPLE II. The closed unit interval, I, is not k-countable.

PROOF. For each n

E q,

let be the set of all open intervals in I having n

i diameter less than

2-

Suppose

[Un

were to have a covering string

Un_

"

Then

,...,

U

}

from 0 to I.

since I is connected, there would be a simple chain

[Unl

nk

That is, 0

Unl,

I U and for each i < i < k

I,

there is a t.

U

O

Un

n

k

n

i+l

But then

i <

Ii

tk_ll

+

Itk_l- tk_21

+.

+

It2

tll +Itll

1 1

+

+

1

+

1

<

2-

+

2nk_l

2n--i i i

This is a contradiction, so that

[n}

cannot have a covering string. Therefore,

by Proposition I0, I is not k-countable. 3

The next three results are consequences of Example

II.

EXAMPLE 12. The Cantor set,

,

is not k-countable.

PROOF. Since there exists a continuous function from ( onto I, then

cannot be k-countable because of Proposition 9 and Example

II.

J

Our next proposition then follows from Example 12 and Proposition 8. PROPOSITION 13. No k-countable space contains a Cantor set.

K-SPACE FUNCTION SPACES 709

PROOF. Let X be k-countable, let x

E

X, and let U be an open neighborhood

of x in X. Since X is completely regular, there exists an f

E

C(X) such that

f(x) 0,

f(X\U)

[I,

and f(X)

=

I. -Snce I is not k-countable by Example ii, and since f(X) is k-countable by Proposition 9, then there exists a t

lf(X).

-i

Thus

[O,t)

N

f(X) is both open and closed in

f(X),

so that f

([0,t))

is an open and closed neighborhood of x contained in U. []

With all these necessary conditions which k-countable spaces must satisfy,

one might wonder whether there exists an uncountable k-countable space. This is

answered by the next two examples.

We will call a space X virtually,

countable

if there exists a finite subset

F of X such that for every open subset U of X with F

=

U, it is true that

X\U

is countable. Notice that a first countable virtually countable space is

c ountable.

PROPOSITION 15. Every virtually countable space is k-countable.

PROOF. Let F be a finite subset of X such that every open U in X with

F

=

U has countable complement, and let

[4n}

be a sequence of open covers for finite subsets of X. First let U_I

41

be such that F

=

U

I.

Then

X\U

I

is countable; say X\U

I

[Xll,

x12, x13,... ].

Let U

2

42

be such that

F

U

[Xll]

=

U

2-

Now

X\U

2 is also countable; say X\U2

Ix21,

x22,

x23,

.

Let U

3 U3 be such that F

U

[Xll,

x12,

x21]

=

U

3-

Continuing by induction,

we may define string

[Un]

from

[ln}

such that for each n, U

n

X\[Xnl,Xn2,

Xn3,

]

and

Fll[Xll,

,Xln,

x21,

,X2,n_

I,

,Xnl=Un+

I.

To see that every

elment

of X is residually in

[Un],

let x

E

X. If x U n=l n

then x is residually in

[Un].

If x

nO__l

U

n,

then let i be the first integer

such that x U.._i Then x x.. for some j, so that for every n i

+

j,

xU

13 n

EXAMPLE 16. The space of ordinals,

,

which are less than or equal to the

first uncountable ordinal is k-countable.

PROOF. It is easy to see that is virtually countable. []

EXAMPLE 17. The Fortissimo space, IF, is k-countable, where IF is I with

the following topology: each

It]

is open for t

# O,

and the open sets containing

0 are the sets containing 0 which have countable complements. Also

]2

is not

Lindelf,

which shows that the converse of Proposition 6 is not true.

PROOF. Obviously IF is virtually countable. However, an alternate proof

can be obtained from known properties of this space. In particular, it follows

from

[I]

that C

()

is homeomorphic to a Z-product of copies of

,

and from

[3]

that a E-product of first countable spaces is a

Frchet

space.

The spaces in the previous two examples are not first countable. This

raises the following question.

QUESTION 18. Is every first countable k-countable space countable?

One well studied example of an uncountable first countable space which is

also a o-dimensional Lindelf space and which does not contain a Cantor set is the Sorgenfrey line. However, in our last example we show that this space is

not k-countable, and in fact is not even

T-countable.

EXAMPLE 19. The Sorgenfrey line, S, is not q-countable. This shows that

the converse of Proposition 5 is not true.

PROOF. For each A

E

(S), let 6(A)

"

min

[la-a

[a,a

E

A, with a

#a

and let U(A)

U[a,a+6(A))la

A}.

Then define U

[U()IA(S),

where

A

Il[-ala

A}.

Clearly I is an open cover for finite subsets of S. Then

{U21UI’

is an open cover of S

2.

But each U

2,

for U

E

l, intersects the set

[(x,y)

S2!x+y

O

on a finite set, so that

[U21U

U}

has no countable sub-cover of S

2.

Therefore no countable subcollection of _{l can cover all doubleton}

K-SPACE FICTION SPACES 711

REFERENCES

i. H. H. Corson, Normality in subsets of product

spaces,

Amer. J. Math. 81

(1959), 785-796.

2. R. A. McCoy,

Countabil,

ity properties, of function spaces, to appear in

Rocky Mountain J. Math.

3. N. Noble, The continuity of functions on Cartesian products, Trans. Amer.